A telescope of length 36 cm forms image of magnification 5 at far point. Focal length of objective and eye piece are

If the telescope is an astronomical telescope , then $L=f_o+f_e=36\mathrm{cm}$ and $\frac{f_o}{f_e}=5$

$\Rightarrow f_o=30\mathrm{cm}\text{, and }{f}_e=6\mathrm{cm}$

If the telescope is a Galelean telescope , then $f_0+f_e=36 .......\left(3\right) and \frac{f_0}{f_e} =- 5 ............\left(4\right)$

Solving $f_o=45 cm and f_e=-9 cm .$